Share & grow the world's knowledge!
We want to connect the people who have knowledge to the people who need it, to bring together people with different perspectives so they can understand each other better, and to empower everyone to share their knowledge.
The midpoints of the sides of triangle ABC are the points D, E and F. Given that the coordinates of D, E and F are (5, 1), (6, 5) and (0, 3), respectively, find the coordinates of A, B and C.
In a triangle, any vertex can be found by adding the two adjacent midpoints and subtracting the opposite midpoint. To find vertex A (opposite D), we calculate A = E + F – D, resulting in (1, 7). To find vertex B (opposite E), we calculate B = D + F – E, resulting in (–1, –1). To find vertex C (opposRead more
In a triangle, any vertex can be found by adding the two adjacent midpoints and subtracting the opposite midpoint. To find vertex A (opposite D), we calculate A = E + F – D, resulting in (1, 7). To find vertex B (opposite E), we calculate B = D + F – E, resulting in (–1, –1). To find vertex C (opposite F), we calculate C = D + E – F, resulting in (11, 3). These are the three vertices.
For Detailed Solutions:
Visit NCERT Solutions for Class 9 Ganita Manjari Chapter 1 Orienting Yourself: The Use of Coordinates Question Answer:
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-1/
See lessA city has two main roads which cross each other at the centre of the city. These two roads are along the North–South (N–S) direction and East–West (E–W) direction. All the other streets of the city run parallel to these roads and are 200 m apart. There are 10 streets in each direction. (i) Using 1 cm = 200 m, draw a model of the city in your notebook. Represent the roads/streets by single lines. (ii) There are street intersections in the model. Each street intersection is formed by two streets — one running in the N–S direction and another in the E–W direction. Each street intersection is referred to in the following manner: If the second street running in the N–S direction and 5th street in the E–W direction meet at some crossing, then we call this street intersection (2, 5). Using this convention, find: (a) how many street intersections can be referred to as (4, 3). (b) how many street intersections can be referred to as (3, 4).
(i) The city is represented as a coordinate grid where North-South streets are vertical lines and East-West streets are horizontal lines. (ii) (a) Based on the convention that an ordered pair represents a specific crossing, only one unique intersection exists for the point (4, 3), where the 4th NortRead more
(i) The city is represented as a coordinate grid where North-South streets are vertical lines and East-West streets are horizontal lines. (ii) (a) Based on the convention that an ordered pair represents a specific crossing, only one unique intersection exists for the point (4, 3), where the 4th North-South street meets the 3rd East-West street. (b) Likewise, there is only one unique intersection referred to as (3, 4). These represent two different physical locations in the city.
For Detailed Solutions:
Visit NCERT Solutions for Class 9 Ganita Manjari Chapter 1 Orienting Yourself: The Use of Coordinates Question Answer:
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-1/
See lessA computer graphics program displays images on a rectangular screen whose coordinate system has the origin at the bottom-left corner. The screen is 800 pixels wide and 600 pixels high. A circular icon of radius 80 pixels is drawn with its centre at the point A (100, 150). Another circular icon of radius 100 pixels is drawn with its centre at the point B (250, 230). Determine: (i) whether any part of either circle lies outside the screen. (ii) whether the two circles intersect each other.
(i) Both circles are within the screen because their centers plus or minus their radii stay between 0 and 800 for width and 0 and 600 for height. (ii) The distance between center A (100, 150) and center B (250, 230) is calculated as 170 pixels. The sum of their radii is 80 plus 100, which equals 180Read more
(i) Both circles are within the screen because their centers plus or minus their radii stay between 0 and 800 for width and 0 and 600 for height. (ii) The distance between center A (100, 150) and center B (250, 230) is calculated as 170 pixels. The sum of their radii is 80 plus 100, which equals 180 pixels. Since the distance between the centers is less than the combined radii, the two circular icons must overlap or intersect.
For Detailed Solutions:
Visit NCERT Solutions for Class 9 Ganita Manjari Chapter 1 Orienting Yourself: The Use of Coordinates Question Answer:
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-1/
See lessPlot the points A (2, 1), B (–1, 2), C (–2, –1) and D (1, –2) in the coordinate plane. Is ABCD a square? Can you explain why? What is the area of this square?
By calculating the distances between the points, we find that all four sides are equal to the square root of 10 units. To confirm it is a square, we check the diagonals: both AC and BD are equal to the square root of 20 units. Since all sides are equal and both diagonals are equal, ABCD is a perfectRead more
By calculating the distances between the points, we find that all four sides are equal to the square root of 10 units. To confirm it is a square, we check the diagonals: both AC and BD are equal to the square root of 20 units. Since all sides are equal and both diagonals are equal, ABCD is a perfect square. The area is calculated by squaring the side length, which gives a total of 10 square units.
For Detailed Solutions:
Visit NCERT Solutions for Class 9 Ganita Manjari Chapter 1 Orienting Yourself: The Use of Coordinates Question Answer:
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-1/
See lessNCERT solutions for class 9 Sanskrit Sharada for 2026-27
NCERT Solutions for Class 9 Sanskrit Sharada (2026–27) provide detailed, chapter-wise answers according to the latest NCERT syllabus and NEP 2020 guidelines. These solutions cover all sixteen chapters with clear explanations of grammar, word meanings and exercises. With simple language and accurateRead more
NCERT Solutions for Class 9 Sanskrit Sharada (2026–27) provide detailed, chapter-wise answers according to the latest NCERT syllabus and NEP 2020 guidelines. These solutions cover all sixteen chapters with clear explanations of grammar, word meanings and exercises. With simple language and accurate solutions, students can achieve better academic performance in new Class 9 Sanskrit Sharada book (संस्कृत शारदा).
Class 9 NCERT/CBSE New book Sanskrit Sharada Chapter List:
Chapter 1 – सत्यं शिवं सुन्दरं संस्कृतम्
Chapter 2 – सुखस्य मूलं धर्म: धर्मस्य मूलम् अर्थ:
Chapter 3 – आत्मवत्सर्वभूतेषु यः पश्यति सः पण्डितः
Chapter 4 – न खलु वयस्तेजसो हेतुः
Chapter 5 – एषा सा कृतकबुद्धि: मानवबुद्ध: सहकरी
Chapter 6 – मनःपूतं समाचरेत
Chapter 7 – उपायं चिन्तयेत प्राज्ञस्तथापायं च चिन्तयेत
Chapter 8 – अन्नाद् आनन्दं प्रति
Chapter 9 – कृतं प्रतिकृतं भूयादेष धर्मः सनातनः
Chapter 10 – णमो अरिहन्ताणम्
Chapter 11 – वर्णोच्चारण-शिक्षा २
Chapter 12 – अन्वयः
Chapter 13 – समासः
Chapter 14 – वाच्यम्
Chapter 15 – शब्दरूपाणि
Chapter 16 – धातुरूपाणि
For more NCERT Solutions of Class 9 Sanskrit Sharada (संस्कृत शारदा) All Chapter Solutions (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/sanskrit/
See less